π does not equal 4
axiomaticdoubt.com
axiomaticdoubt.com
Imagine a circle and its diameter. Now imagine two circles with half of the diameter, lined up so that the diameter lines align. Now split those two in four, etc. The circle becomes a snaking line whose total length doesn't change, and the snaking line converges uniformly to the line. Clearly, however, pi is not 1.
What you need is convergence in position _and_ angle. A curve that converges in position and angle _does_ converge in length: the reference I know which shows this is reference [9] on the above-mentioned paper. (edit: in case you don't want to download the gigantic file --- yay for publishing in graphicsy places --- the reference is: K. Hildebrandt, K. Polthier, and M. Wardetzky. On the convergence of metric and geometric properties of polyhedral surfaces. Geometriae Dediacata, (123):89–112, 2006.)
Under uniform convergence, the limit of the integrals is equal to the integral of the limit. So, this works fine for areas, or are you talking about something else?
I was trying to say that convergence in area for two-dimensional surfaces in R^3 requires convergence in normals in the same way that convergence in length for one-dimensional curves in R^2 requires convergence in normals.
For area in R^2, volume in R^3, and so on, you're definitely right.
One interesting and useless law that this reveals is that the perimiter of any convex blob of pixels is equal to the perimiter of its bounding rectangle.
It's easy enough with arguments like this to show the "length" to be anything you choose. Such demonstrations are instructive.
EDIT: Changed lower-case ell to upper-case for clarity.
Wikipedia has a nice short article about the ℓ₁ norm (also called “taxicab distance” or “Manhattan distance”): http://en.wikipedia.org/wiki/Taxicab_geometry
It explains,
> “The use of Manhattan distance leads to a strange concept: when the resolution of the Taxicab geometry is made larger, approaching infinity (the size of division of the axis approaches 0), it seems intuitive that the Manhattan distance would approach the Euclidean metric [...] but it does not. This is essentially a consequence of being forced to adhere to single-axis movement: when following the Manhattan metric, one cannot move diagonally (in more than one axis simultaneously).”
A troll can also prove that 2 == 1 by continuously folding the peaks of an equilateral triangle down to the baseline.
Another fun one:
$1 = 100¢
$.1 = 10¢
$.1^2 = 10¢ ^ 2
$.01 = 100¢
thus
$1 = 1¢
Trollface
Understanding you to be a distinguished algebraist (that is, distinguished from other algebraists by different face, different height, etc.), I beg to submit to you a difficulty which distresses me much.
If x and y are each equal to 1, it is plain that
2 * (x^2 - y^2) = 0, and also that 5 * (x - y) = 0.
Hence 2 * (x^2 - y^2) = 5 * (x - y).
Now divide each side of this equation by (x - y).
Then 2 * (x + y) = 5.
But (x + y) = (1 + 1), i.e. = 2. So that 2 * 2 = 5.
Ever since this painful fact has been forced upon me, I have not slept more than 8 hours a night, and have not been able to eat more than 3 meals a day.
I trust you will pity me and will kindly explain the difficulty to Your obliged,
Lewis Carroll.
You can't divide by 0 you just get nonsense.
The problem really is that you can't divide by zero, even in an algebraic expression.
A simpler example of this phenomenon (which blew my mind when I first encountered it) occurs with the equation x = x^2. If you divide by x, you get x = 1, which is a solution to the equation, but where did the other solution x = 0 go??
Whenever you divide an equation by an algebraic expression, you need to consider the possibility of that expression being zero and treat it as a special case. So in the case of x = x^2, you can reason as follows: maybe x = 0, in which case … what … ah yes, that's a solution! Or maybe x ≠ 0, in which case we can divide by it and get x = 1. That doesn't contradict the assumption x ≠ 0, so it's okay, and x = 1 is the other solution.
So, the actual problem is dividing by zero. Your assumption that "we are dealing with algebra and not numerical values" is false because it completely ignores the "if x and y = 1" part.
0.01 (dollars)^2 = 100 (cents)^2
0.01 (dollars)^2 = 100 (1/100 dollars)^2
Real analysis was very satisfying, in somewhat the same way that building low-level software or libraries is satisfying -- I got to understand the guts. It was also fun to learn about erroneous historical assumptions made due to insufficient rigor. IIRC, until at least the 1870s, it was believed that any continuous function must be differentiable almost everywhere, and that in fact this should be obvious. It turns out that one can construct continuous functions that are nowhere differentiable!
This was one of my favorite topics in the class:
http://en.wikipedia.org/wiki/Construction_of_the_real_number...
That is exactly the reason why taking a course in real analysis made me switch majors. I was trying to get that feeling from physics, but only found it to my satisfaction in math.
"Just because a series of curves tends to a limit curve doesn't mean that the series of lengths of those curves tends to the length of the limit curve" is the whole of the explanation.
If we write down area and perimeter error e_A and e_P as a function of n, in the limit of infinite n e_A is proportional to 1/n. However, e_P is independent of n, since the perimeter does not change at any step. Thus, if the perimeter was unequal to begin (clear by inspection), it does not get closer through this approximation.
The mistake is in accepting that anything that looks like a circle must be a circle. The fractal beast has area arbitrarily close to that of a circle, but it's clear once we look at e_A and e_P that it is not an approximation of its perimeter.
Pointwise, the limit is a circle. Every point on the enclosing shape gets mapped to a sequence of points. Each of these sequences has a limit, that limit is on the circle. The resulting implied mapping of the original square to the circle is a continuous bijection. By every sense that we usually talk of the convergence of lines, this enclosing shape does approach the circle in the limit.
The point is that the mapping involved is not a length-preserving mapping.
The circle is continuously differentiable along its circumference. The fractal is not---it's still got a large number of discontinuities (offhand, it looks like the number of discontinuities is equal to 2^(n-1) after n steps).
The area enclosed by this curve approaches that of a circle, but the curve is not a circle: it cannot be described by x^2+y^2=1 because its derivatives are not equal to those of same.
The circumference of the limit doesn't equal the limit of the circumferences.
We can have a bijection from the original square to the circle. Why is this a surprise? It's not length preserving. We can have a bijection from the interval [0,1] to the interval [0,2]. That's not length preserving either.
If I've misunderstood you then perhaps you could explain your thinking in more detail. I don't understand where you think there are two versions of a circle.
The usual picture given to non-mathematicians about sequences and limits can end up being strongly misleading in cases like this. Just because a sequence never "gets there," the limit is still the limit. This is the same kind of murky area that talks about 0.999... recurring never "getting to" 1. It doesn't have to "get to" 1 because it's never travelling.
It's also the kind of problems that arise when talking about proof by induction. Talking about dominoes falling down is, in the longer term, very misleading. We prove P(1), and we prove that P(n) => P(n+1), then they are all true. They don't become true one by one, they are simply all true - it is what it is.
I hope that helps.
I'm thinking of starting a blog to talk about things like this - it falls between the levels of the non-mathematician and true researcher.
Edit: Nevermind, reading cousin_it's posting I think I've got a handle on it. My confusion was exactly the difference between a sequence of approximations, and the limit itself.
From the puzzle, name the square object S and the circle inside it C_1.
Imagine another circle C_2 that circumscribes the square S from the puzzle. i.e. the corners of the square lie on the circle C_2. Then for each step, when we invert the outer-most corners of the square, we constrict C_2 such that the circle lies on the new outer-most points of S.
What happens is that as you repeat this process more and more the outer circle C_2 gets smaller and smaller, approaching the size of the original circle inside the square, C_1.
Also you can infer that the area of S is equal to the area of C_1 and C_2 since (area C_2) -> (area C_1) and (area C_1) <= (area S) <= (area C_2). Which makes sense intuitively, too, since they all enclose the same space.
This tells you nothing about the relationships between the circumference of the objects, though.
Read the post closely. The author introduces a set S of squarey curves approximating a circle. This set has an obvious correspondence with the natural numbers: there is curve #1, curve #2, etc. Then the author defines a function f: S->S that takes curve #n to curve #n+1. Wait, that doesn't sound right! The function f has no interesting structure whatsoever, it's exactly equivalent to defining f(n)=n+1 on the naturals. Of course, taking the "limit" of a function f: S->S makes no sense at all.
What would make sense is taking the limit of a certain function N->C, where N is the naturals and C is the set of all curves on the plane. That is, the limit of a sequence of curves (not of a function from curves to curves as the OP tried to say). To talk about such limits, you need to define what it means for a sequence of curves to converge - a "topology" on C. There are many ways to do that, some more outlandish than others. One way is pointwise convergence: assume a parameterization t->C_i(t) on each curves in the sequence, and require that C_i(t) converges co C_lim(t) for each t separately.
Now, pretty much any reasonable notion of "convergence" on the space of all curves has to imply pointwise convergence. That is, pointwise convergence is a very "weak" notion of convergence: if we have a sequence of curves that has a limit in some reasonable sense, then it had better coincide with the pointwise limit, dontcha think?
And here we come to the second facepalm moment in the post. Under pointwise convergence, the sequence of squarey curves under discussion does not converge to some "right angled fractal beast". It converges to the circle. As n grows, every point on S_n comes closer and closer to some point on the circle. Ain't nothing more to it.
Now the correct explanation for the original puzzle. Pointwise convergence of curves doesn't imply that their lengths converge to the limit's length. Hell, we don't even need 2-dimensional space to show that! A simple straight line will do. Imagine a human traveling a straight road of 1km length in this fashion: he takes two steps forward, then one step back, then repeats. In the end he will have traveled about 3km instead of 1km. As we make the human and his steps tinier and tinier, his movement looks smoother and smoother to an external observer, but he still travels 3km in total instead of 1. Or maybe (going back to the 2D space) the human could take a step left, then forward, then right; this would make his path look like a fine comb that approximates the straight line more and more closely, but it's always 3x longer. Something like this is happening in the original puzzle.
Finishing touch: there are notions of convergence where it's true that the length of the curves in the sequence always converges to the length of the limit. One such notion says that the direction of travel (velocity vector) must also pointwise converge to the velocity vector of the limit curve. Under this definition of convergence, the original sequence of curves does not converge, because it makes too many sharp turns.
I'm pretty uneducated about math, so please let me know what trap I'm falling into.
Your answer isn't yet helping me because you just note that Manhattan Distance remains the same no matter how many twists and turns there are. Yes, this is the premise of the question!
But turn up the twistiness to infinity. Now the Manhattan Distance line is identical to the hypotenuse. There is no point on the Manhattan distance line isn't also on the hypotenuse and vice versa. They both have an infinite number of points, of the same aleph-number, I think.
You mention a hypothetical path that doubled back on itself -- it is easy to see why that would result in a different answer. But the Manhattan Distance line is not doubling back on itself. Every point is a step towards the goal, and it doesn't cross itself. All of its deviations from the simple hypotenuse are infinitely small.
So why are the distances still different?
In this case, it's mathematically simple that a Manhattan line should have the same length, no matter what N (the number of steps) is.
[1] http://en.wikipedia.org/wiki/Russell%27s_paradox for instance.
The Banach-Tarski paradox is also worthwhile: http://en.wikipedia.org/wiki/Banach%E2%80%93Tarski_paradox
I think nowadays most mathematicians are "comfortable" with such aberrant results, but that was not always the case. And there's still a chance that future generations of mathematicians will dismiss this work.
The number of points isn't relevant - the Cantor Set (also known as Cantor Dust) is an uncountably large subset of the real line, and yet its length is zero.
I guess the only insight I can offer at this point is that length has nothing to do with points.
The side of a regular polygon is √(sin²(τ/n)+(1−cos(τ/n))²)=√(2(1−cos(τ/n))), while the corresponding arc is τ/n.
d/dn τ/n−√(2(1−cos(τ/n))) = −(τ/n²+sin(τ/n)/√(2(1−cos(τ/n)))) < 0, so the difference is strictly decreasing (note that sin(τ/n)>0 when n>2).
So there's your lower bound for the difference. You'd still have to prove the limit is at 0, though.
(This is the "velocity vector" convergence condition mentioned by cousin_it above).
No they won't. It moves zero distance horizontally and zero distance vertically. All the points of the infinitely zigzagging line are on the hypotenuse.
Of course, now we're back to Zeno; if it's not moving off the line, and it only makes progress when it's off the line, how can it get anywhere?
You can't talk about moving "zero distance horizontally and zero distance vertically" becuase then, as you rightly say, we're back to Zeno. You don't have an infinitely zigzagging line. That way of trying to think about things is a dead end, and unhelpful.
Better answers are given elsewhere in this thread.
The true line has an uncountable infinity of points on the line, and the difference between an uncountable infinity and a countable one is an uncountable infinity, so we're still way the hell off.
+ The limit has all the points on the line.
+ Uncountable sets can be of length zero
+ There are families of curves where at no stage do any of the points lie on the circle, and yet the limit is the circle, and the limit of the length is the length of the circle.
In short, your comment isn't really explaining anything.
ADDED IN EDIT: Whoa - cool - a downvote! Please, let me know where I'm wrong. I thought I'd explained clearly why the comment was wrong, and I have explained elsewhere more about what's really going on. No doubt you, the downvoter, have moved on now, but whoever you are, I'd love more information as to what you think I've got wrong.
Thanks.
Here, let's try: the square is 1x1. Consider one quarter of the circle: radius of the circle is 1/2, half a diagonal of the square is sqrt(2) / 2, diagonal of the removed square is (rt(2) - 1) / 2. Area of the square removed is ((rt(2) - 1) / 2) ^ 2. (This is trivial via the pythagorean theorem, saving some math.)
Alright, that's the first square we accumulate. Now the magic happens: every step, we cut the square's side in half, but make two of them. Agree with me so far? Good. If we cut the side of a square in half, we cut its area to a quarter, but since we have two squares now the total area is 1/2 of the last square. Agree with me so far? Good. We can trivially sum infinite geometric series: t1 / (1 - r), where t1 is the first term and r is the fraction each term gets multiplied by. In this case, it turns out that in any one quadrant the sum of the series of squares removed is 2 * ((rt(2) - 1) / 2) ^ 2, or just (rt(2) - 1)^2 / 2.
Multiply by 4 to get the picture over all four quadrants, and we get 2 * (rt(2) - 1) ^ 2. A little simplification and we get 2 * (2 - 2 rt(2) + 1) = 6 - 4 rt(2)
So, we've got a unit square, so the area of the square is 1. If we subtract the area of the infinite series of squares, we get 4 rt(2) - 5 =~ .657. We expect the area of the circle to be pi / 4 =~ 0.7853975. Thus, the square minus and infinite series of squares doesn't approximate the known area of a circle at all.
The problem lies not with the area, but the perimeter; the figure's area turns out to be unrelated for this problem.
Thus, argumentation from area will not work here.
The paradox lies in "infinity" and "never". Achilles will overtake the tortoise when he's one atom away from the tortoise, and similarly at one point your corner removal will reach the atom level, where you can no longer reduce it and maintain a square shape.
At atom level, when your squares consist of three atoms in a L pattern, you can't reduce it further without distorting the squares
Assuming, of course, that atoms are the smallest particles.
4 - π will be that small, tiny difference.
I meant that difference that you no longer see (when you go to infinite), not 0,85.